Answer:
1.84 kJ (kilojoules)
Explanation:
A specific heat of 0.46 J/g Cº means that it takes 0.46 Joules of energy to raise the temperature of 1 gram of iron by 1 Cº.
If we want to heat 50 g of iron from 20° C to 100° C, we can make the following calculation:
Heat = (specific heat)*(mass)*(temp change)
Heat = (0.46 J/g Cº)*(50g)*(100° C - 20° C)
[Note how the units cancel to yield just Joules]
Heat = 1840 Joules, or 1.84 kJ
[Note that the number is positive: Energy is added to the system. If we used cold iron to cool 50g of 100° C water, the temperature change would be (Final - Initial) or (20° C - 100° C). The number is -1.84 kJ: the negative means heat was removed from the system (the iron).
The Euglena is unique in that it is both heterotrophic (must consume food) and autotrophic (can make its own food).
Answer:
Oh well have your parents do something bout that. They are there to help you. Or guardian they all shall help you that is what they are there for. They are they to take care of you, answer you questions, and last but not least, they are there to also raise you. Ask you mom, dad, guardian, and maybe even you teacher to post pone the date. Teachers are to help you. Your guardian is to watch you make sure your safe while your parents are away (bascially like raising you). And parents to raise you, make sure you safe, and many other things like to make sure you are fed.
Explanation:
I hope they all have some type of help for you.
Answer:
Explanation:
λ=c x²
c = λ / x²
λ is mass / length
so its dimensional formula is ML⁻¹
x is length so its dimensional formula is L
c = λ / x²
= ML⁻¹ / L²
= ML⁻³
B )
We shall find out the mass of the rod with the help of given expression of mass per unit length and equate it with given mass that is M
The mass in the rod is symmetrically distributed on both side of middle point.
we consider a small strip of rod of length dx at x distance away from middle point
its mass dm = λdx = cx² dx
By integrating it from -L to +L we can calculate mass of whole rod , that is
M = ∫cx² dx
= [c x³ / 3] from -L/2 to +L/2
= c/3 [ L³/8 + L³/8]
M = c L³/12
c = 12 M L⁻³
C ) Moment of inertia of rod
∫dmx²
= ∫λdxx²
= ∫cx²dxx²
= ∫cx⁴dx
= c x⁵ / 5 from - L/2 to L/2
= c / 5 ( L⁵/ 32 +L⁵/ 32)
= (2c / 160)L⁵
= (c / 80) L⁵
= (12 M L⁻³/80)L⁵
= 3/20 ML²
=
=
